9.29 [Cr(NH3)6] 3+ is paramagnetic while [Ni(CN)4] 2– is diamagnetic. Explain why?

Overall charge balance in [Cr (NH3)6]3+ complex: X + 6 (0) = + 3 X = + 3 Cr is in + 3 oxidation state. Electronic configuration of Cr in + 2 state: 3d3 . Now ammonia is a weak field ligand so it not causes pairing of the unpaired electron and undergoes hybridisation to form 6 sp3d2 hybrid orbitals filled by the six ammonia ligands. It's geometry is octahedral with unpaired electrons and hence is paramagnetic complex. In case of [Ni (CN)4]2– ion : Overall charge balance in [Ni (CN)4]2–complex: X + 4 (-1) = -2 X = + 2 Ni is in + 2 oxidation state. Electronic configuration of Ni in + 2 state: 3d8. Now cyanide ion is a strong field ligand so it causes pairing of the unpaired electron and undergoes hybridisation to form 4 dsp2 hybrid orbitals filled by the four cyanide ligands. It's geometry is square planar with no unpaired electrons and hence is diamagnetic complex.

9.29 [Cr(NH₃)₆] ³⁺ is paramagnetic while [Ni(CN)₄]^2– is diamagnetic. Explain why?

4 Views|Posted 10 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

10 months ago

Overall charge balance in [Cr (NH₃)₆]³⁺ complex:

X + 6 (0) = + 3 X = + 3

Cr is in + 3 oxidation state.

Electronic configuration of Cr in + 2 state: 3d³ . Now ammonia is a weak field ligand so it not causes pairing of the unpaired electron and undergoes hybridisation to form 6 sp³d² hybrid orbitals fill